# Ideal class “group” of Lipschitz (fully-integer) quaternions

Let $H = \left\{a+bi+cj+dk \in \mathbb{H} : a,b,c,d \in \mathbb{Z} \;\mbox{ or }\, a,b,c,d \in \mathbb{Z} + \tfrac{1}{2}\right\}$ be Hurwitz (or semi-integer) quaternions. Then $H$ is Euclidean, thus principal ideal domain.

Now let $L = \left\{a+bi+cj+dk \in \mathbb{H} : a,b,c,d \in \mathbb{Z}\right\}$ be Lipschitz (or fully-integer) quaternions. It is easy to see that the right ideal $(1+i+j+k, 2)$ is not principal. How can I prove using $H$ that the ideal class "group" $Cl(L)$ contains only two elements, the classes of $(1)$ and $(1+i+j+k,2)$? Here, the ideal class "group" $Cl(L)$ is a set of non-zero right ideals $I \subset H$ modulo the relation $I_1 \sim I_2 \Leftrightarrow x_1I_1=x_2I_2$ for $x_1, x_2 \in L$.

• You need to distinguish between one-sided and two-sided ideals. In the one-sided case you don't get a group, but you do in the 2-sided case. From your definition of ideal classes, it looks like you are either looking at right ideals or two-sided ideals. – Kimball Oct 12 '16 at 12:02
• I am missing something. $2=(1-i)(1+i)=(1+i)(1-i)$, so $2$ is an element of both the left and right ideal generated by $(1+i)$. IOW that ideal is principal, no? – Jyrki Lahtonen Oct 22 '16 at 22:50
• Kimball, Jyrki, thank you, I corrected the question. – evgeny Oct 30 '16 at 6:19

## 1 Answer

There is a direct solution using modified Euclidean algorithm. The problem is that this algorithm does not always make the norm smaller. Anyway, we can prove directly that each right ideal is either $(\alpha)$ or $(\alpha, \alpha\tau)$ for $\tau=\frac{1+i+j+k}{2}$.

Consider $\alpha, \beta \in L$ and approximate $\beta^{-1} \alpha$ by an element $q \in A$ so that the difference $\left(\beta^{-1} \alpha-q\right)$ has all coordinates with respect to $\{ 1, i, j, k \}$ in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right]$. If there is a coordinate not equal to $\frac{1}{2}$, then $\alpha=\beta q+r$ and $N(r)<N(\beta)$. Otherwise the difference is $\tau=\frac{1+i+j+k}{2}$, so $\alpha=\beta q+\beta \tau$ and $N(\beta)=N(\beta \tau)$.

Now consider a right ideal $I \triangleleft L$ and fix a finite number of its generators. If two generators has different norms, then divide them (if the quotient is zero, forget it), so the sum of norms is now smaller. If the norms are equal, but the division make the norm of the divident smaller, also do it. This way after a finite number of steps we get a finite set of generators with equal norms, such that the division of any two of them does not reduce the norm. So, if one of them is $\alpha$, all other may be replaced by $\alpha\tau$.